Question: Simplify the following expression: $y = \dfrac{-3x^2+13x- 12}{-3x + 4}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-3)}{(-12)} &=& 36 \\ {a} + {b} &=& &=& {13} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $36$ and add them together. The factors that add up to ${13}$ will be your ${a}$ and ${b}$ When ${a}$ is ${4}$ and ${b}$ is ${9}$ $ \begin{eqnarray} {ab} &=& ({4})({9}) &=& 36 \\ {a} + {b} &=& {4} + {9} &=& 13 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-3}x^2 +{4}x) + ({9}x {-12}) $ Factor out the common factors: $ x(-3x + 4) - 3(-3x + 4)$ Now factor out $(-3x + 4)$ $ (-3x + 4)(x - 3)$ The original expression can therefore be written: $ \dfrac{(-3x + 4)(x - 3)}{-3x + 4}$ We are dividing by $-3x + 4$ , so $-3x + 4 \neq 0$ Therefore, $x \neq \frac{4}{3}$ This leaves us with $x - 3; x \neq \frac{4}{3}$.